Entity Manager nicht korrekt durch die Feder eingespritzt und gibt java.lang.NullPointerException

Question

Ich bin ein Neuling bei Frameworks in JAVA EE mit so, wenn ich versuche, eine Methode des Unternehmens-Manager zu nennen es wirft ein java.lang.NullPointerException so ich will, was ich schrieb dies ist ein Beispiel für meine dao Implementierung

@Repository 
@Transactional 
public class IUserDAOImpl implements IUserDAO{ 

    @PersistenceContext 
    private EntityManager em; 

    public void addUser(User u) { 
     em.persist(u); 
    } 
} 

das ist mein Service

@Service 
public class IUserServiceImpl implements IUserService { 

@Autowired 
private IUserDAO dao = new IUserDAOImpl(); 

public void setDao(IUserDAO dao) { 
    this.dao = dao; 
} 

public void addUser(User u) { 
    dao.addUser(u);  
} 

das ist meine Aktionsklasse

public class ConnectionAction extends ActionSupport { 

User c = new User("a","b","c",11,"d",true); 

@Autowired 
public IUserService service; 


public String execute() throws Exception { 
    service.addUser(c); 
     return SUCCESS; 
} 

die web.xml

<?xml version="1.0" encoding="UTF-8"?> 
<web-app id="struts_blank" version="2.4" 
xmlns="http://java.sun.com/xml/ns/j2ee" 
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" 
xsi:schemaLocation="http://java.sun.com/xml/ns/j2ee   http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd"> 
<display-name>Struts Blank</display-name> 

<filter> 
<filter-name>struts2</filter-name> 
<filter-class> 
org.apache.struts2.dispatcher.ng.filter.StrutsPrepareAndExecuteFilter 
</filter-class> 
</filter> 
<filter-mapping> 
<filter-name>struts2</filter-name> 
<url-pattern>/*</url-pattern> 
</filter-mapping> 

<context-param> 
<param-name>contextConfiguration</param-name> 
<param-value>classpath:applicationContext.xml</param-value> 
</context-param> 

<listener> 
<listener-class> 
org.springframework.web.context.ContextLoaderListener 
</listener-class> 
</listener> 
</web-app> 

die applicationContext.xml (obwohl es wirklich so vermasselt, wenn Sie sometihng falsch bemerkt erklären PLZ)

<?xml version="1.0" encoding="UTF-8"?> 

<beans xmlns:tx="http://www.springframework.org/schema/tx"  xmlns="http://www.springframework.org/schema/beans"  xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" 
xmlns:context="http://www.springframework.org/schema/context"  xmlns:util="http://www.springframework.org/schema/util" 
xsi:schemaLocation=" 
http://www.springframework.org/schema/beans   http://www.springframework.org/schema/beans/spring-beans-2.5.xsd 
http://www.springframework.org/schema/context  http://www.springframework.org/schema/context/spring-context-2.5.xsd 
http://www.springframework.org/schema/util  http://www.springframework.org/schema/util/spring-util-2.0.xsd 
http://www.springframework.org/schema/tx  http://www.springframework.org/schema/tx/spring-tx-2.5.xsd"> 

<bean id="dao" 
class="com.iticsys.GBO.dao.IUserDAOImpl"> 
</bean> 

<bean id="service" 
class="com.iticsys.GBO.services.IUserServiceImpl"> 
<property name="dao" ref="dao"></property> 
</bean> 

<bean id="dataSource" class="org.springframework.jdbc.datasource.DriverManagerDataSource"> 
<property name="driverClassName" value="com.mysql.jdbc.Driver" /> 
<property name="url" value="jdbc:mysql://localhost:3306/GBO1" /> 
<property name="username" value="root" /> 
<property name="password" value="" /> 
</bean> 

<bean id="persistenceUnitManager" 
    class="org.springframework.orm.jpa.persistenceunit.DefaultPersistenceUnitManager"> 
<property name="persistenceXmlLocations"> 
<value>classpath*:META-INF/persistence.xml</value> 
</property> 
<property name="defaultDataSource" ref="dataSource"> </property> 
</bean> 

<bean id="entityManagerFactory" 
class="org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean"> 
<property name="persistenceUnitManager" ref="persistenceUnitManager"></property> 
<property name="persistenceXmlLocation" value="classpath*:META-INF/persistence.xml" /> 
<property name="persistenceUnitName" value="persistenceUnit" /> 
</bean> 

<bean id="transactionManager" class="org.springframework.orm.jpa.JpaTransactionManager"> 
<property name="entityManagerFactory" ref="entityManagerFactory"></property> 
</bean> 
<tx:annotation-driven transaction-manager="transactionManager" /> 
<context:annotation-config></context:annotation-config> 
<context:component-scan base-package="com.iticsys.GBO.dao" /> 
</beans> 

persistence.xml

<persistence xmlns="http://java.sun.com/xml/ns/persistence" 
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" 
xsi:schemaLocation="http://java.sun.com/xml/ns/persistence http://java.sun.com/xml/ns/persistence/persistence_2_0.xsd" 
version="2.0"> 
<persistence-unit name="persistenceUnit" 
    transaction-type="RESOURCE_LOCAL"> 
    <provider>org.hibernate.jpa.HibernatePersistenceProvider</provider> 


    <properties> 
     <property name="javax.persistence.jdbc.driver" value="com.mysql.jdbc.Driver" /> 
     <property name="javax.persistence.jdbc.url" value="jdbc:mysql://localhost:3306/GBO1" /> 
     <property name="javax.persistence.jdbc.user" value="root" /> 
     <property name="javax.persistence.jdbc.password" value="" /> 

     <property name="hibernate.show_sql" value="true" /> 
     <property name="hibernate.hbm2ddl.auto" value="update" /> 
    </properties> 
</persistence-unit> 

und vielen Dank im Voraus

Answer 1

Bei der Implementierung der Klasse IUserDAOImpl sollten Sie das Objekt von EntityManager wie

private EntityManager em = new EntityManager(/*constructor parameter if any*/);

oder Sie können eine Referenz von Objekt von EntityManager instanziiert zuweisen, Sie können in Ihren anderen Unterklassen wie IUserServiceImpl, ConnectionAction erstellen.

Dafür können Sie eine Methode in Ihrer Klasse hinzufügen IUserDAOImpl die diese reference.like erhalten,

public void setEntityManager(EntityManager em_ref){ em = em_ref;} 

Answer 2

public class HibernateUtil { 

static Logger logger = Logger.getLogger(HibernateUtil.class); 

private static final String FAILED_TO_CREATE_SESSION_FACTORY_OBJECT = "Failed to create sessionFactory object."; 
private static final SessionFactory sessionFactory = buildSessionFactory(); 
private static ServiceRegistry serviceRegistry; 

private static SessionFactory buildSessionFactory() { 

try { 
    Configuration configuration = new Configuration(); 
    configuration.configure("./hibernate.cfg.xml"); 

    configuration.addAnnotatedClass(Calzz.class); 

    serviceRegistry = new StandardServiceRegistryBuilder().applySettings(configuration.getProperties()).build(); 

    return configuration.buildSessionFactory(serviceRegistry); 

} catch (Throwable ex) { 

    logger.debug(FAILED_TO_CREATE_SESSION_FACTORY_OBJECT + ex); 
    throw new ExceptionInInitializerError(ex); 
} 
} 

public static SessionFactory getSessionFactory() { 
return sessionFactory; 
} 

}